Linear Graphs (Cambridge (CIE) O Level Additional Maths): Revision Note

• Quickly sketching the graph of the straight line(s) can be helpful if you are struggling with an exam question

• Ensure you state equations of straight lines in the format required

  • Usually  $y=mx+c$  or  $ax+by+d=0$

  • Check whether coefficients need to be integers (they usually are for $ax+by+d=0$)

(a) Find the equation of the straight line with gradient 3 that passes through (5, 4).

We know that the gradient is 3 so the line takes the form 

$y=3x+c$

To find the value of c, substitute (5, 4) into the equation

$\begin{array}{l}4=3\left(5\right)+c\\ 4=15+c\\ c=-11\end{array}$

Replace c with −11 to complete the equation of the line

y = 3x − 11

(b) Find the equation of the straight line that passes through (-2, 6) and (8, 1).

You may find it helpful to sketch the information given

First find m, the gradient

$\begin{array}{rcl}m& =& \frac{6-1}{-2-8}\\ & =& \frac{5}{-10}\\ & =& -\frac{1}{2}\end{array}$

We know that the line takes the form 

$\begin{array}{rcl}y& =& -\frac{1}{2}x+c\end{array}$

To find the value of c, substitute either of the given points into this equation. Here we will pick (8, 1) as it is doesn't contain negative numbers so is easier to work with

$\begin{array}{l}1=-\frac{1}{2}\left(8\right)+c\\ 1=-4+c\\ c=5\end{array}$

Replace c with −11 to complete the equation of the line

$\mathit{y}\mathbf{=}\mathbf{-}\frac{\mathbf{1}}{\mathbf{2}}\mathit{x}\mathbf{+}\mathbf{5}$

We can check against our sketch that this equation looks correct- it has a negative gradient and it crosses the y-axis between 1 and 6

The coordinates of A are (−4, 3) and the coordinates of B are (8, −12).

Find M, the midpoint of AB.

The midpoint can be found using M = $\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right)$

Fill in the values of x and y  from each coordinate

$\left(\frac{-4+8}{2},\frac{3+-12}{2}\right)=\left(\frac{4}{2},\frac{-9}{2}\right)$

Simplify

M = (2, −4.5)

• Work with the square of a distance for as long as possible as this avoids early rounding errors

  • In the non-calculator paper your answer may need to be left as a surd

• Only square root when forced to or for a final answer, and use the ANS button (and other memory features) on your calculator

Point A has coordinates (3, -4) and point B has coordinates (-5, 2).

Calculate the distance of the line segment AB.

Using the formula for the distance between two points, $d=\sqrt{{\left(x_1-x_2\right)}^2+{\left(y_1-y_2\right)}^2}$ 

Substituting in the two given coordinates:

$d=\sqrt{{\left(3--5\right)}^2+{\left(-4-2\right)}^2}$

Simplify: 

$d=\sqrt{{\left(8\right)}^2+{\left(-6\right)}^2}=\sqrt{64+36}=\sqrt{100}=10$

Answer = 10 units